wanminliu@gmail.com
§
/
6.5
Surfaces
and
surfaces
ds
=/
Fix
IF
/
dudv
integrals
=1¥÷i+i+l¥
an
•
what
is
a
surface
?
-
parametric
surface
in
IR
>
Write
ri=2÷
.
ñ=a÷
One
Define
R=
[
a.
b)
✗
[
c.
d
]
CIR
'
piece
(
f
:
R
→
IR
>
(
Uw
)
↳
(
✗
luau
)
.
ylu.rs
,
2-
LUND
}
=EEHñ%tñ-F
r
:*
.
"
{
glue
deferent
pieces
⇐
C
,
1=2
CIRZ
composite
surface
•
surface
integral
lrixrii-lriifuis-ino-lriilriictu.io
)
§
f-
IX.
Y
.
2-
)
IS
=/
riilrii-lru.mil
'
wanminliu@gmail.com
We
use
the
variable
0
instead
of
t
.
#
I
(a)
Verify
that
on
the
plane
curve
r
=
910
)
di
=
(
✗
'
(
o
)
do
)
't
4^101
do
)
'
with
polar
equation
the
an
length
element
=
⑨
'
lo
)
Coot
gloss
-
since
)
)
do
]
"
is
given
by
1-
[
(910151^0+910)
Gso
)
do
]
"
ds=§ioD
'
do
=
¢910
)
)
't
⑨
'
copy
do
'
(b)
What
is
the
area
element
on
the
vertical
cylinder
given
in
terms
of
cylindrical
coordinates
50
Is
=f9lO)i
'
do
by
r
=
g
(
o
)
(b)
✗
z
coordinate
fr
,
O
,
Z
)
dS=
dsdz
SI
(a)
if
the
plane
curve
is
parametrized
=
§ioy=§io)#
today
,
y
,
,
,
,
,
an
,
,
,
,
,
,
,
in
any
,
,
,
an
µgi,
,
coor dinat e
,
then
✗
ds
'
=
Exits
dtj-fy.lt/dt5
Now
✗
=
r
cos
0
=
910
)
G)
O
Y
=
r
Sino
=
9101
Sino
wanminliu@gmail.com
#
7
Find
ffxds
over
the
part
of
The
two
cylinders
internet
a
curve
in
the
5
1-
St
octant
.
the
parabolic
cylinder
2-
=
¥
2-
=
¥
Fix
,=
(
×
,
.
É
)
that
lies
inside
the
first
octant
part
{
✗
45=1
/
✗
zo
Of
X
E
1
of
the
cylinder
✗
45=1
.
I
470
If
X=o
:
10
,
I
,
0
)
A
✗
=L
:
(
1
,
0
,
£
)
13
The
sinful
5
is
on
the
parabolic
cylinder
2-
=
¥
,
but
is
bounded
by
curve
Fix
)
the
curve
013
,
and
the
curve
0A
.
✗
"
÷÷÷÷÷÷÷÷T
The
projection
of
the
surface
5
to
the
Xy
-
plane
>
Y
is
the
region
D=
(
§
of
the
unit
disc
)O
CA
[
with
C
=
(
1.
0,0
)
±
wanminliu@gmail.com
Computation
of
double
integral
Recall
for
a
graph
2-
=
2-
ix.
y
)
4.
7)
c-
D
f)
✗
tot
dxdy
ExAMPLE4-
D
with
D=
d
IX.
y
)
ftp.d/XZ-y2E1
,
✗
70
.
430
}
☒
fix
.
>
.
2-
ids
=
f)
fcx.y.2-ix.yDHZ.it#dxdy
☐
IS
✗
txt
dxd
>
D
=
fi
dxf
.
dy
f
✗
txt
)
ffxds
=
fSxFÉzj
dxdy
5
☐
=
g.in#i-xTdx/Z=-E,Zx--X.2-y--
0
=
If
'
txt
DX
'
)
✗
I
t
c-
10,1
]
✓
=
I
1
!
ktyi-#
dt
=
If
!
dt
XY
D
=
IS
✗
tÑd×
"
÷
g.zwso.ws
•
do
*
°
"
"
1-
=
Sino
,
Otto
.
¥3
✗
=
rcoso
A
y
=
r
Sino
recoil
]
=
§
IF
ws2÷
do
=
-14
t-iz .fi? cos 20d 20-
0
C-
[
0
,
I
]
0
c
'
✗
=
F-
.
g-
lids
-
mm
%¥
,
=
0
.
wanminliu@gmail.com
Method
so
we
still
have
%"×
D
✗
:*
.
i
±
Y
-
cylinder
"
¥n
ds
=
dsidy-i-xfdxdyds-rdx-E-i-2-ixTdx-i-xt.de
the
are
length
on
the
X
-
Z
-
plane
for
the
curve
2-
=
¥
.
☐
wanminliu@gmail.com
Let
D
be
the
pr ojection
of
5
onto
XY
-
plane
.
#
13
Find
IS
yds
.
5
Hy
=
zc×4yT
✗
'
+
=
,
where
5
is
the
part
of
plane
2=1+7
1+5+2
>
=
"
"
"
"
|
D
↳
the
ref
"
"
that
lies
inside
the
cone
2-
=2li
.
2×45-4=1
it
4¥
£1
.
2×2-1
-
1)
'
=
2
Sol
.
The
plane
and
the
cone
On
the
plane
2-
=
it
y
,
then
internet
at
a
curve
dS=Hzj
dxdy
Morning
:
{
Z
=
icy
=<1+d
the
surf ace
is
on
the
plane
2-
=
2(X¥y
So
we
need
the
=D
dxdy
plane
equation
here
.
So
2-
30
.
1+770
Y
→
-1
It
is
NO T
the
cone
2-
IT
equation
i
IS
yds
=
If
y
Jzdxdy
2=21×7%1
I
5
D
>
y
=
22
9
€
We
give
two
methods
for
computation
in
next
page
.
wanminliu@gmail.com
Computation
.
=
252
/
'
Cristo
rad s
q
-
.
I
=
rzffydxdy
y
-1
D
F
=
S
C-
FI
.
I
]
dy=
Rds
0
since
odd
function
D=
{
ix.
y
)
/
✗
4
El
}
y=
Is
-11
#
ds
+
4rsH
✗
Hits
=
4
Lis
-_
sin
0
.
Off
I
]
=
2K
s¥do=s¥i=i¥
.•a!
=
+
zisinyl.I.i.in
;
>
✗
recon
,
=
-÷
Of
Co
,
27
]
Y¥=r
Sino
ydyf
"¥~
5-
if
ffydxdy
=
V2
f
"
"
DX
Jacobian
=
1¥
¥
=/
"
0
"
"
"
"
/
=
far
☐
us
_tH
¥
¥
.
/
rzsino
arioso
5-
if
do
/
Ikr
)
drlrzrsino
-11
)
µ
fz
Y.tt#TdY
=
2
so
"dofirdr+zrYdfdmdr
1-
if
=
2.
27
-
I
=
2k
.
☐
wanminliu@gmail.com
https://wanminliu.github.io/doc/DG/DG.html
②
A
level
surface
is
orientable
.
§
16.6
Oriented
integrals
and
let
f
:
IR
}
→
IR
be
a
smooth
function
flux
integrals
and
t
EIR
a
regular
value
.
An
oriented
surface
is
a
smooth
S
=
f-
"
CH
is
a
regwarsmtag-yeyn.mn
For
any
p
c-
S
surface
together
with
a
choice
of
global
✓
Pflp
)
=
(
¥
,
(
p
)
,
(
pi
,
¥+1M
)
-1-10.99
unit
normal
vector
field
Ñcp
)
.
Then
gp
,
¥
0S
is
an
orientation
.
Example
latent
①
A
graph
2-
=
f-
ix.
y
)
is
orientable
③
The
Miibious
strip
is
NOT
orientable
.
With
Lecture
8
.
Ñcp
,
=
tt×,-f
-
suppose
there
is
a
normal
"
"
tradition
,
Hf×Ztf
direction
at
A.
we
mov e
it
along
the
red
curve
After
one
"
circle
"
the
normal
vector
goes
to
the
opposite
direction
,
a
contradiction
.
So
there
is
No
global
normal
Vertov
field
wanminliu@gmail.com
#
5
Find
the
flux
of
got
.
Methodist
%Ñ
✗
✗
=
-
zx
¥
=
✗
it
yJ+zÉ
2
>
=
-
ay
upward
through
the
part
of
the
surface
Ñ
=
f-
2-
×
,
-
2-
y
,
1)
☐
HZ×4ZÑ
2-
=
a
-
✗
2-
y
'
=
124m€
lying
above
the
plane
2-
=b
,
where
b<
a
.
HzE+zÑ
Question
:
Why
this
Ñ
is
upward
?
Answ er
:
Because
its
third
component
is
POSITIVE
.
Flux
of
vector
field
¥
acros s
theory
surface
5
É=
ix.
y
.
t
)
I
.
Ñ
=
If
É
•
Ñds
=
ff-F.DE
dS=Hhi+Ñdxdy
S
5
When
the
surface
is
closed
,
the
flux
IS
F.
Ñds
=
§
(2×7-277-2)
dxdy
5
→
on
surfaces
=
Sf
(
✗
4-
Y'
+
a)
dxdy
=
I
integral
is
devoted
by
#
I.
di
we
put
the
2-
=
a-
✗
Zyl
D
where
D
is
the
disc
D=
{
IX.
Y
>
/
✗
'
+
y
'
E
la
-
b)
}
wanminliu@gmail.com
Method
computation
:
d§=ÑdS=-¥ ¥F*Id×b
D=
{
IX.
y
>
/
✗
'
+
y
'
E
la
-
b)
}
I
=
a.
Area
(D)
+
{
§
Cxiyydxdy
SO
d§
=
C-
2-
×
,
-
2-
>
a)
dx dy
rEE-iyzdxdy-rdrdorc-CO.AT
]
,
O
C-
[
0,27
]
=
a.
ala
-
b)
+
fi→r3drfoy
Since
☒
=
(
X
,
Y
,
Z
)
.
4-
b)
2
.
27
4-
É
.
d§=
-
✗
2×-44
+
I
¥
Yay
a-
X
-
y
-
=
a
(
a'
-
ab
+
É
+
É
-
ab
)
=
X'
+
y'
+
a
.
=
a
(
ta
'
-
Zab
-1¥
)
f)
É
.
di
=
fflxtiixajdxdy
.
5
D
we
then
compute
as
method
1
.
☐
wanminliu@gmail.com
✗
z
#
9
Find
the
flux
of
=
✗
I
+
yj
✗
Y
upward
through
the
part
of
the
surface
,
.
.
2-
=
2-
X'
-
2yd
>
7
j
>
✗
-
-1
that
lies
above
the
xy
-
plane
.
I
D=
/
Him
/
✗
Fry
-
⇐
z
}
Sol
.
If
I.
D8
I.
Is
>
=
2/11×7-25
)
dxdy
J
D
d§
=
C-
zx
,
-
2-
y
,
1)
dx dy
please
I
2521T
.
compute
it
d§
=
(
ZX
,4Y
,
1)
dxdy
by
yourself
.
Hint
:
You
may
use
your
pr evious
☐
É
.
I
}
=
(
✗
,
y
,
0
)
(
211,49
,
1)
dxdy
computation
in
Question
#
8
IS
1×4
>
4d×b=%¥I
É
=
(2×2+45)
dxdy
☐
Ipnta-b
wanminliu@gmail.com
#
'
}
Find
the
flux
of
Remark
.
By
Examp1e2
E-
7¥
,
out
of
the
surface
of
the
cube
É
=
☐
of
-
as
x.y.es
a
(
with
a
>
9
.
with
10=0/4
.
y
,
2)
=
-1¥
Sol
.
let
we
denote
the
surtout
☒
É
.
D8
=
☒
f)
D8
s
s
cube
by
5
,
which
is
piecewise
smooth
wth
6
pieces
You
could
read
§
17.4
,
Example
4
.
⑧
S
=
Sit
£+5
}
-15×+5++56
.
☐
⑦
E.
di
can
be
computed
by
5
Ae
decomposition
⑧
L
wanminliu@gmail.com
#
15
Define
the
flux
of
a
plane
vector
field
acr oss
a
piecewise
smooth
curve
.
Find
the
flux
of
(a)
✗
=
awso
Y
=
a
Sino
0
C-
10
.
27
]
É=
xityj
we
use
0
instead
of
s
outward
arcos
,
(a)
the
circle
XYY
!
a
"
(
with
a
>
9
For
any
law
>
o
,
a
shot
on
the
circle
(b)
the
boundary
of
the
sfuare
-1
EX
,
yet
.
A
=
@
so
,
sino
)
§
.
Ñ
=
✗
woo
+
y
since
=
a
define
it
as
f
E.
ñds
e
{
ads
=
a.
(27-9)
=
22
a
?
wanminliu@gmail.com
(b)
{
É
.
Ñds
C
14 . 11
11
.lt
B
=
f
É
.
Ña
,
ds
-1
f
E.
Ñpeds
D
(-1-1)
11
,
-
1)
A
ÑB
BT
+
⇐
¥
-
Ñaods
+
§gÉ
-
rinds
^
F-
=
IX.
4)
Nap
=
(
1.0
)
Now
ÑBC
=
(
Q1
)
f
=p
.
Ña☐
ds
=
f
✗
ds
=
[
I
1.
dy
=
2
AÑ
ÑB
Ñcp
=
C-
1,0
)
0h
AI
✗
=/
,
with
S
=
Y
,
-1
EYE
/
ÑDA
=
(
0
,
-
1)
Similarly
for
other
integrals
.
So
{
E.
ñds
=
2×4
=
8
☐
